SOLUTION: Just not getting this. Suppose cos t = -0.4 and csc t < 0 . Find each of the following sin t = tan t = csc t = sec t = cot t = I know cos = -4/10 = -2/5 and cs

Algebra ->  Trigonometry-basics -> SOLUTION: Just not getting this. Suppose cos t = -0.4 and csc t < 0 . Find each of the following sin t = tan t = csc t = sec t = cot t = I know cos = -4/10 = -2/5 and cs      Log On


   

Question 921046: Just not getting this.
Suppose cos t = -0.4 and csc t < 0 . Find each of the following
sin t =
tan t =
csc t =
sec t =
cot t =
I know cos = -4/10 = -2/5
and csc = 1/y and this states that csc<0 so I'd think that is saying y must be negative.
cos = adj/hyp so I use Pythagorean theorem and since y is negative I know it must be in either III or IV quadrant but which is negative out of cos the 2 or 5? I think it's the 5. Which I go to figure out the opposite I run into this problem:
b = √(-5^2 + 2^2)
b = √(-25+4)
b = √(-21)
But I can't square a negative?
What am I misunderstanding here?
I know the answer for b is -√21/5 I just don't know how to get that
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose cos t = -0.4 and csc t < 0 . Find each of the following
sin t =
tan t =
csc t =
sec t =
cot t =
***
Given data shows reference angle t is in quadrant III in which cos<0, sin<0
cos t=-0.4=-4/10=-2/5
sin t=-√(1-cos^2t)=-√(1-4/25)=-√(21/25)=-√21/5
tan t=sin/cos=-√21/-2=√21/2
sec t=-5/2
csc t=-5/√21=-5√21/21